GATE Papers >> CSE >> 2007 >> Question No 69

Question No. 69 CSE | GATE 2007

The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used, is:

Answer : (B) $\left\lceil\log_2\frac{2LtR}K\right\rceil$

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